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Given,

$\ {y}_{i} = N({\mu}_{i}, {\Sigma }_{i}) $

If we go by the link http://www.tina-vision.net/docs/memos/2003-003.pdf then we can understand that the product of many multivariate gaussians can be written as:

$ \prod {y}_{i} = {y}_{p} = N({\mu }_{p}, {\Sigma }_{p})$

Where,

$\Sigma_{p}^{-1} = \sum \Sigma_{i}^{-1}$

and $\Sigma_{p}^{-1}{\mu }_{p} = \sum \Sigma_{i}^{-1}{\mu }_{i}$

What can we say about the product $ \prod {Y}_{i}$ of gaussian processes given by:

$\ {Y}_{i} = GP({m}_{i}\left(x \right),{k}_{i}\left(x,x' \right))$

kjetil b halvorsen
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Ankit Chiplunkar
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    Hi, welcome to the site. The linked document talks about multiplying Gaussian probability density functions while your notation in the questions sounds like you are asking about products of Gaussian random variables (which will not be Gaussian). 1. Do you understand the difference? 2. Which one are you asking about? – Juho Kokkala Dec 18 '14 at 12:18
  • @JuhoKokkala I agree that the document talks about multiplying probability density functions (PDF's)

    1. I also see that there is a difference between PDF and Gaussian Random variable. I found another link here that talks about Gaussian random variable's being multiplied link, and their product being a Gaussian

    2. I want to know if we can find the mean and covariance function of product of GP's by extrapolating the information about product of multivariate gaussians?

     – Ankit Chiplunkar Dec 18 '14 at 13:40
  • @JuhoKokkala I am pretty new to statistics.
    Am I using the two terms very loosely? In my understanding A gaussian random variable (GRV) is a random variable with PDF resembling a gaussian function. So, when I talk about multiplying two GRV's and their product being a Gaussian, I mean to say that the PDF of product of these two GRV's will also be resemble a gaussian function. Please correct me if I am wrong.     – Ankit Chiplunkar Dec 18 '14 at 13:49
  • @AnkitChiplunkar, you are wrong. The PDF of a product of two gaussian r.v.s will not be a product of two Gaussain PDFs. It's very easy to see why. – Aksakal Dec 18 '14 at 13:58

2 Answers2

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Let two independent normal (Gaussian) random variables,

$$X \sim N(\mu_x, \sigma^2_x),\;\;\; Y \sim N(\mu_y, \sigma^2_y)$$

with probability density functions $f_X(x)$ and $f_Y(y)$ respectively. Then the probability density function of the product of the two random variables, i.e. of the random variable $Z = XY$ is

$$ f_Z(z) = \int^{\infty}_{-\infty} f_X \left( x \right) f_Y \left( z/x \right) \frac{1}{|x|}\, dx$$

As you can see, the density of $Z$, $f_Z(z)$, is not the product of the densities. Informally, this is because the probaility density function does not determine what values the $Z$ variable takes, but how probability is allocated to the values that $Z$ takes (values that are determined by some other function, usually unspecified).

Alecos Papadopoulos
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  • Thanks a lot, I understood. So, taking product of two PDF's f(x), g(y) will result in their joint proabbility distribution. – Ankit Chiplunkar Dec 18 '14 at 14:23
  • Yes, if they are independent. – Alecos Papadopoulos Dec 18 '14 at 14:32
  • To clarify, $f_Z(z)$ will not simply be the product of the underlying Gaussian processes with probability density functions $f_X(x)$ and $f_Y(y)$. But cannot $f_Z(z)$ still be a Gaussian process itself (albeit with a possibly more complicated mean and covariance function)? – Mathews24 Mar 06 '19 at 18:10
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    @Mathews24 The original question falls for a fatal confusion between the product of two random variables, and the product of the PDF's of two random variables. I would suggest to you also to stop using the PDF's as notational representations of the random variables themselves. A random variable is a function, and distinct from its PDF, which is another function. $Z=XY$ cannot be a Gaussian random variable. – Alecos Papadopoulos Mar 06 '19 at 19:45
  • @AlecosPapadopoulos It's the proof that $Z = XY$ cannot be a Gaussian random variable that I am missing. I agree that $f_Z(z)$ is not simply the product of the densities. But I am not quite seeing how this proves that $Z$ is not a normally distributed variable. Is this based upon a separate proof? – Mathews24 Mar 08 '19 at 06:27
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    @Mathews24 An existential (dis)proof would be along the lines of 'the Normal distribution is not cosed under multiplication". But it suffices to derive the pdf of $Z$ and observe that it is not and it cannot be the same as the pdf of a Normal random variable. The product of two independent Normals is a studied distribution, see for example, http://mathworld.wolfram.com/NormalProductDistribution.html and Craig (1936). – Alecos Papadopoulos Mar 08 '19 at 08:29
  • Fair point, here is a simple example for possibly interested future readers. – Mathews24 Mar 08 '19 at 16:59
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The product of Gaussian processes will not be a Gaussian process, unlike the sum of Gaussian processes. When you multiply to random variables, you don't simply multiply their PDFs, it's very easy to see why.

Aksakal
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